Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int_"0"^pi (x"d"x)/(1 + sin x)`
Advertisements
उत्तर
Let I = `int_"0"^pi (x"d"x)/(1 + sin x)` .....(i)
= `int_0^pi (pi - x)/(1 + sin(pi - x)) "d"x` ......`["Using" int_0^"a" "f"(x) "d"x = int_0^"a" "f"("a" - x)"d"x]`
= `int_0^pi (pi - x)/(1 + sinx) "d"x` ......(ii)
Adding (i) and (ii), we get
2I = `int_0^pi (x/(1 + sinx) + (pi - x)/(1 + sinx)) "d"x`
= `int_0^pi ((x + pi - x)/(1 + sinx))"d"x`
= `int_0^pi pi/(1 + sin x) "d"x`
= `pi int_0^pi 1/(1 + sinx) "d"x`
= `pi int_0^pi (1.(1 - sinx))/((1 + sinx)(1 - sinx)) "d"x`
= `pi int_0^pi (1 - sinx)/(1 - sin^2x) "d"x`
= `pi int_0^pi (1 - sinx)/(cos^x) "d"x`
= `pi int_0^pi (1/(cos^2x) - sinx/(cos^2x))"d"x`
= `pi int_0^pi (sec^2x - secx tanx)"d"x`
= `pi[tanx - sec]_0^pi`
= `pi[tan pi - tan 0) - (sec pi - sec 0)]`
2I = `pi[0 - (-1 - 1)`
= `pi`(2)
∴ I = `pi`
Hence, I = `pi`
APPEARS IN
संबंधित प्रश्न
Integrate the rational function:
`(3x + 5)/(x^3 - x^2 - x + 1)`
Integrate the rational function:
`(2x - 3)/((x^2 -1)(2x + 3))`
Integrate the rational function:
`(x^3 + x + 1)/(x^2 -1)`
Integrate the following w.r.t. x : `x^2/((x^2 + 1)(x^2 - 2)(x^2 + 3))`
Integrate the following w.r.t. x : `(2x)/(4 - 3x - x^2)`
Integrate the following w.r.t. x:
`(6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1)`
Integrate the following w.r.t. x : `2^x/(4^x - 3 * 2^x - 4`
Integrate the following w.r.t. x: `(2x^2 - 1)/(x^4 + 9x^2 + 20)`
Integrate the following with respect to the respective variable : `(cos 7x - cos8x)/(1 + 2 cos 5x)`
Integrate the following w.r.t.x : `x^2/sqrt(1 - x^6)`
Integrate the following w.r.t.x : `(1)/((1 - cos4x)(3 - cot2x)`
Integrate the following w.r.t.x : `(1)/(2cosx + 3sinx)`
Evaluate:
`int (2x + 1)/(x(x - 1)(x - 4)) dx`.
State whether the following statement is True or False.
If `int (("x - 1") "dx")/(("x + 1")("x - 2"))` = A log |x + 1| + B log |x - 2| + c, then A + B = 1.
Evaluate: `int ("3x" - 1)/("2x"^2 - "x" - 1)` dx
Evaluate: `int (2"x"^3 - 3"x"^2 - 9"x" + 1)/("2x"^2 - "x" - 10)` dx
`int "e"^(3logx) (x^4 + 1)^(-1) "d"x`
`int ((x^2 + 2))/(x^2 + 1) "a"^(x + tan^(-1_x)) "d"x`
`int 1/(4x^2 - 20x + 17) "d"x`
`int sec^2x sqrt(tan^2x + tanx - 7) "d"x`
`int "e"^(sin^(-1_x))[(x + sqrt(1 - x^2))/sqrt(1 - x^2)] "d"x`
`int (3x + 4)/sqrt(2x^2 + 2x + 1) "d"x`
`int xcos^3x "d"x`
`int (3"e"^(2x) + 5)/(4"e"^(2x) - 5) "d"x`
Verify the following using the concept of integration as an antiderivative
`int (x^3"d"x)/(x + 1) = x - x^2/2 + x^3/3 - log|x + 1| + "C"`
If f(x) = `int(3x - 1)x(x + 1)(18x^11 + 15x^10 - 10x^9)^(1/6)dx`, where f(0) = 0, is in the form of `((18x^α + 15x^β - 10x^γ)^δ)/θ`, then (3α + 4β + 5γ + 6δ + 7θ) is ______. (Where δ is a rational number in its simplest form)
Evaluate: `int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx`
Evaluate`int(5x^2-6x+3)/(2x-3)dx`
Evaluate.
`int (5x^2 - 6x + 3) / (2x -3) dx`
Evaluate:
`int(2x^3 - 1)/(x^4 + x)dx`
If \[\int\frac{2x+3}{(x-1)(x^{2}+1)}\mathrm{d}x\] = \[=\log_{e}\left\{(x-1)^{\frac{5}{2}}\left(x^{2}+1\right)^{a}\right\}-\frac{1}{2}\tan^{-1}x+\mathrm{A}\] where A is an arbitrary constant, then the value of a is
Integration by partial fractions is a method used to integrate which functions?
Which partial form corresponds to \[\frac{\mathrm{p}x^{2}+\mathrm{q}x+\mathrm{r}}{(x-\mathrm{a})(x-\mathrm{b})(x-\mathrm{c})}\]?
Why is \[\frac{x^{2}+1}{x^{2}-5x+6}\] not a proper rational function?
